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    <p>Your problem strongly reminds me of the calculations that have to be done for <a href="http://en.wikipedia.org/wiki/Motion_compensation#Block_motion_compensation" rel="nofollow noreferrer">motion compensation</a> in video compression. Maybe you should take a closer look what's done in that area.</p> <p>As rlbond already pointed out, counting the number of points in a window where the colors exactly match isn't what's normally done in comparing pictures. A conceptually simpler method than using discrete cosine or wavelet transformations is to add the squares of the differences</p> <pre><code>diff = (m_color[x1 + xWindow][y1 + yWindow] - m_color[x2 + xWindow][y2 + yWindow]); sum += diff*diff; </code></pre> <p>and use <em>sum</em> instead of <em>match</em> as criterion for similarity (now smaller means better). </p> <p>Back to what you really asked: I think it is possible to cut down the running time by the factor 2/sizeWindow (maybe squared?), but it is a little bit messy. It's based on the fact that certain pairs of squares you compare stay almost the same when incrementing y1 by 1. If the offsets xOff = x2-x1 and yOff = y2-y1 are the same, only the top (rsp. bottom) vertical stripes of the squares are no longer (rsp. now, but not before) matched. If you keep the values you calculate for match in a two-dimensional array indexed by the offsets xOff = x2-x1 and yOff = y2-y1, then can calculate the new value for match[xOff][yOff] for y1 increased by 1 and x1 staying the same by 2*sizeWindow comparisons:</p> <pre><code>for (int x = x1; x &lt; x1 + sizeWindow; x++) { if (m_color[x][y1] == m_color[x + xOff][y1 + yOff]) { match[xOff][yOff]--; // top stripes no longer compared } if (m_color[x][y1+sizeWindow] == m_color[x + xOff][y1 + sizeWindow + yOff]) { match[xOff][yOff]++; // bottom stripe compared not, but wasn't before } } </code></pre> <p>(as the possible values for yOff changed - by incrementing y1 - from the interval [y2 - y1, k_maxY - sizeWindow - y1 - 1] to the interval [y2 - y1 - 1, k_maxY - sizeWindow - y1 - 2] you can discard the matches with second index yOff = k_maxY - sizeWindow - y1 - 1 and have to calculate the matches with second index yOff = y2 - y1 - 1 differently). Maybe you can also keep the values by how much you increase/decrease match[][] during the loop in an array to get another 2/sizeWindow speed-up.</p>
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    1. COThank you so much!
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